5 The following diagram shows a simplified model for an auto
Solution
solution:
1) here input from support is x(t)=Xsinwt
x\'(t)=Xwcoswt
x\'\'(t)=-Xw2sinwt
and it will produce periodic output
y(t)=Ysinwt
where
y\'(t)=Ywcoswt
y\'\'(t)=-Yw2sinwt
2) here absolute deflection of spring is =(y-x)
absolute velocity=(y\'-x\')
where FBD of mass has spring force,damping force ,inertia acting vertically upward,hence equation of motion is
my\'\'+cy\'+ky=cx\'+kx
putting value in equation we get
my\'\'+cy\'+ky=X(ksinwt+ccoswt)
as term in bracket is in triangle law hence we can write
my\'\'+cy\'+ky=X(k2+c2w2)^.5*(cosasinwt+sinacoswt)
where
cosa=k/(k2+c2w2)^.5 and sina=cw/(k2+c2w2)^.5
hence equation is
my\'\'+cy\'+ky=X(k2+c2w2)^.5*(sinwt+a)
this equation is similar to
mx\'\'+cx\'+kx\'=F0sinwt
where F0=X(k2+c2w2)^.5
where solution is given by
y=yc+yp
yc=Y(exp^-(zeta*wn*t))sin(wdt+phi)
where yp=Ysin(wt+a-phi)
3) hence on solving steady state amplitude of vibration iss given by
Y=X*((k2+c2w2)^.5/k)/(((1-(w/wn)2)2+(2*zeta*(w/wn))2)^.5)
4) where transfer function which is ratio of output y(t) to input x(t) is given as
H(t)=y(t)/x(t)
H(t)=Ysinwt/Xsinwt
H(t)=Y/X=((k2+c2w2)^.5/k)/(((1-(w/wn)2)2+(2*zeta*(w/wn))2)^.5)
5)to simulate system let consider w=wn,zeta=0 low value,and input amplitude ofX= .1 m
so we get
Y=infinite
hence for low value of damping vehicla produce to unpleasant ride and unpleasant ride turn pleasant with addition of damper in the sytsem
6) system produce pleasant ride if damping factor is within underdamped condition and value of spring stiffness k should be high enough, mass M should be low and road condition has to be good means w of road input should be high enough means its amplitude should be as small as possible
hence zeta<1,
K as high as possible,
M=low
w=low frequency of disturbance
good ride means when relative amplitude od mass is zero,that is
z(t)=y(t)-x(t)=0

